3 1 IIO 1 III 2 3 4 3 1 2 3 4 5 1. Is this trait dominant or recessive? Explain your answer. 5 6 7 2. What gave you the essential information to decide that II-3 and II-4 were heterozygous? 678 8 4. Draw a pedigree (info above) showing all four grandparents, the two parents, and the son. Indicate each individual's possible genotypes. 9 3. Brown eyes are a dominant eye-color allele and blue eyes are recessive. A brown-eyed woman whose father had blue eye and whose mother had brown eyes marries a brown-eyed man whose parents are also brown-eyed. They have a son who is blue- eyed.
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- A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?re ||| E 6. Label the genotypes for this pedigree of an X-linked recessive disorder (red- green colorblindness). (a) 2 2 3 1 3 a. How do you know? b. Label the genotypes. 4 O To 2 4 5 5 6 6 7. Is the following pedigree autosomal recessive, autosomal dominant or X-linked recessive? 2 8 O T 58
- zto.mheducation.com/ext/map/index.html?_con%=con&external_browser=0&launchUrl=https%253A%252F%252Flms.mheducation.com%252Fmghmiddleware? er 10 Assignment Saved Classify the following conditions based on whether they are describing autosomal dominance, autosomal recessive, or both. Autosomal Dominant Affected children can have unaffected pped parents Book Print Heterozygotes are affected erences Autosomal Recessive Heterozygotes have a normal phenotype Both males and females are affected with equal frequency Both Affected children have at least one affected parent 080 acer -> %24 % 2. 6.SQ5 Two people are considering having children, but they each have family members that died of a rare recessive autosomal disorder. For the man, it was his brother and for the women, it was her mother’s sister (her aunt). What is the probability they will have an affected child?BIU A- == 三E 12 - Match each of the following examples to the appropriate type of non-Mendelian inheritance. 1. A homozygous recessive genotype for the gene that encodes phenylalanine hydroxylase (which breaks down the amino acid phenylalanine) causes lighter skin color, a musty odor, differences in intellectual development, and seizures. 2. In pea plants, alleles of Gene W control flower color, with the dominant allele (W) leading to purple flower.color, and the recessive allele (w) leading to white flower color. Usually, a genotype of WW or Ww leads to purple flowers. However, when Gene C is homozygous recessive, WW or Ww plants always have white flowers. 3. In mallard ducks, feather coloring is controlled by Gene F. A dominant allele (F) leads to green head feathers, while a recessive allele (f) leads to brown head feathers. In male mallards, inheritance of one or more F alleles always leads to the green head feather trait. But female mallards always have brown head feathers,…
- What type of heredity is shown in the pedigree? (hint: check your notes - "modes of inheritance") 3 Autosomal Dominant Sex-Linked Recessive Autosomal Recessive Sex-Linked Dominant Music off Zoom in 99 esc 26 & 1 2 3 4 5 6 7 8 detete Q w R T. Y P tab A S F K retu caps lock C V B N M shift command option control option command .. ..Please answer all of them, they are all connected. PEDIGREE ANALYSIS and SYMBOLOGY: Examine the pedigree which has X linked Dominant inheritance of disorder. Use letter X* (asterisk denotes disorder) as genotype of the individuals which can be XX, XY, X*X*, X*X and X*Y. a. What is the genotype of IV-6? b. What is the genotype of III-6? c. What is the genotype of II-3? d. What is the genotype of III-8? e. If couple I-1 and I-2 will have a son, what is the probability of having the disorder? f. If couple III-8 and III-9 will have another child, what is the probability of having the disorder? g. Theoretically, if individual IV-3 and individual IV-5 will marry and will have a child, what is the probability of having a child without the X-linked disorder?三|三 主 主 13 ||L 4 6. 8. Construct a Punnett square for a cross between a heterozygous black guinea pig and a homozygous white guinea pig. a. What genotypes would you expect in the offspring? 5.
- 188IS Add-oIS al text BIUA E E E E Times New. 12 1 4. Purple fur (P) is DOMINANT in monsters. Yellow fur (p) is RECESSIVE. What is the genotype of a PURE PURPLE monster? = What is the GENOTYPE of a HETEROZYGOUS purple monster? What is the GENOTYPE of a YELLOW monster? Using the punnet square above, make a cross between TWO HETEROZYGOUS PURPLE MONSTERS. POSSIBLE OFFSPRING GENOTYPES PHENOTYPES edu gear Elage An extra finger in humans is rare but is due to a dominant gene. When one parent is normal and the other parent has an extra finger but is heterozygous for the trait, what is the probability that the first child will have an extra finger? - /mod/quiz/attempt.php?attempt=1173673&cmid=3837312&page=11# Select one: O a. 25% O b. 50% O c. 75% O d. 0% O e. 100% @ 2 W # 3 E с $ 4 R G Search or type URL % 5 T MacBook Pro 6 Y & 7 ☆ U * 8 ( 9 F ) 0 0 Next page P (3) BO 11! I 7 l 8 / Using the pedigree you have constructed, complete the following Punnett square using A/a. 1) Show the cross between Braxton and his non-albino wife who had an Mother's Alleles albino father: Mother's genotype: Father's genotype: Genotype %: Phenotype %: Click to add speaker notes Father's Alleles